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The residue theorem states that a meromorphic function with finitely many poles inside a positively oriented simple closed contour and none on it satisfies ${\oint }_{C}F(z)dz=2\pi i\sum \mathrm{Res}(F;a)$. The function must be holomorphic on a neighbourhood of the contour and its interior away from those poles.

Write $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$ and $I={\int }_0^{\infty }{x}^{1/2}/(1+x^2)dx$. Both integrals converge absolutely. Use the principal complex logarithm in the upper half-plane, with $z^{1/2}=e^{\log z/2}$, and an upper semicircle of radius $R>1$ indented above zero by a clockwise semicircle of radius $ϵ<1$. The branch values on the negative real side are upper limits; equivalently use contours arbitrarily slightly above that side before taking a limit.

For $F(z)=z^{1/2}\log z/(1+z^2)$, the large arc is $O(R^{-1/2}\log R)$ and the small arc is $O({ϵ}^{3/2}|\log ϵ|)$, so both vanish. On $z=-x$ approached from above, $z^{1/2}=i\sqrt{x}$ and $\log z=\log x+i\pi$. The oriented negative segment therefore contributes $iJ-\pi I$, and the positive segment contributes $J$. The only enclosed pole is $i$, with residueConsequentlyComparing imaginary and real parts yields the upper-half-plane contour for square-root logarithmic integrals evaluation:

For $z\ne 0$, the principal value of complex exponentiation is $z^w=\exp (w\mathrm{Log}z)$, where the principal complex logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z\le \pi$. De Moivre's theorem states that $(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$ for every integer $n$.

Since $\sqrt{3}+i=2e^{i\pi /6}$, the six roots are

Use the principal complex logarithm, for which $\mathrm{Log}i=i\pi /2$, and setThen the definition of complex exponentiation givesand thereforeOther choices of branch of the complex logarithm produce further valid answers.

Write $z=r{e}^{i\theta }$ with $r>0$. On a fixed branch of the complex logarithm on which $\log z=\log r+i\theta$ and $\log \bar{z}=\log r-i\theta$, the equation becomesIts real and imaginary parts give the same condition $\log r=\theta$, orThis is a logarithmic spiral. For the principal complex logarithm, it is the portion parametrized by $-\pi <\theta <\pi$ that avoids the branch cut.

The principal cube root is $z^{1/3}=\exp (\mathrm{Log}z/3)$ using the principal complex logarithm. In the right half-plane, its complex argument lies between $-\pi /6$ and $\pi /6$. Consequently the roots of $r^3=-z$ consist of one root with negative real part, $-z^{1/3}$, and two with positive real part, $z^{1/3}{e}^{\pm i\pi /3}$.

Set $I={\int }_0^{\infty }{x}^{1/2}/(1+x^2)dx$ and $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$. In an indented upper semicircle use the principal complex logarithm and $z^{1/2}=e^{\log z/2}$. The negative real boundary contributes $iJ-\pi I$, the positive boundary contributes $J$, and both circular arcs vanish. The pole at $i$ has residue $\pi e^{i\pi /4}/4$. Thus $(1+i)J-\pi I=({\pi }^2/(2\sqrt{2}))(-1+i)$; comparing real and imaginary parts yields the displayed integrals.